# I Can I use the Schrodinger picture when the Hamiltonian is time-dependent?

#### Albfey

In the Schrodinger picture, the operators don't change with the time, but the states do. So, what happen if my hamiltonian depend on time? Should I use the others pictures in these cases?

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#### vanhees71

Science Advisor
Gold Member
What should happen? You can (in principle) use any picture of time evolution you like for any Hamiltonian. If the Hamiltonian is explicitly time dependent, the states still evolve with the operator $\hat{C}(t,t_0)$, which is defined by the operator equation
$$\mathrm{i} \hbar \partial_t \hat{C}=\hat{H}(t) \hat{C}(t,t_0), \quad \hat{C}(t_0+0^+,t_0)=\hat{1}.$$
The formal solution is given by
$$\hat{C}(t,t_0)=\mathcal{T} \exp \left [-\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t' \hat{H}(t') \right].$$
Here $\mathcal{T}$ denotes the time-ordering operator. For a derivation and explanation of the meaning of time-ordering see the first section of

• Albfey and atyy

#### Albfey

So, can I say that, in the Schrodinger picture, every operator is time independent, except (maybe) the hamiltonian? In this case, if I try to compute the expection value of the hamiltonian $\hat{H}(t)$ in the state $| \psi (t) \rangle$, I get $$\langle \psi (t) | \hat{H}(t) | \psi(t) \rangle = \langle \psi (t_{0}) | \hat{C}^{\dagger}(t,t_{0}) \hat{H}(t) \hat{C}(t,t_{0})|\psi(t_{0}) \rangle = \langle \psi (t_{0}) | \hat{H}(t) | \psi(t_{0}) \rangle,$$ where I have used the fact that the hamiltonian comute with the time evolution operator, that is unitary. But isn't this the expection value of a operator in the Heisemberg picture?

#### vanhees71

Science Advisor
Gold Member
In the Schrödinger picture all time dependence of operators that represent observables are by definition explict time dependence. The full time dependence is carried by the state, represented by the statistical operator (for pure states the statistical operator is the projection operator $|\psi(t) \rangle \langle \psi(t)|$).

I don't understand, how you come to the conclusion in the last step. I think it's wrong in general since $\hat{C}(t,t_0)$ and $\hat{H}(t)$ do not commute in general.

• Albfey

#### Albfey

I think it's wrong in general since $\hat{C}(t,t_0)$ and $\hat{H}(t)$ do not commute in general.
Yes, you are right, they don't comute in general. I confused it with another situation, I'm sorry.

In the Schrödinger picture all time dependence of operators that represent observables are by definition explict time dependence.
I have already read it in a book, but I didn't understand what it means. The most books say that all operators are time-independent in this picture. So, how can they have an explicit dependence on time?

I'm sorry for insisting on this question, I am really confused about pictures in quantum mechanics.

#### kith

Science Advisor
An explicit time-dependence in the Hamiltonian means that there's some time-dependency which you can't get rid of by transforming from the Heisenberg picture to the Schrödigner picture. This implies that energy isn't conserved so it can only occur if the quantum system of interest isn't isolated.

A very common example would be the treatment of the interaction of an atom with an external light field by perturbation theory, which involves the semi-classical approximation (i.e. the electric field is not treated as a quantum system but enters the Hamiltonian of the atom as a time-dependent classical field). There, we have transitions between different energy levels.

[Note that in most cases, open systems cannot be modeled simply by making the (Schrödinger picture) Hamiltonian time-dependent. This is because interactions with something external typically lead to entanglement which manifests as a loss of coherence in the system state. If the system interacts with a much bigger environment, this loss is for all practical purposes permanent ("decoherence").]

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• • vanhees71, PeterDonis and Albfey

#### Albfey

Thank you, guys. I was really confused about pictures. I have read more about this and now I think this is clearer, and your answers helped me. Sorry for the slightly confused question.

#### vanhees71

Science Advisor
Gold Member
Yes, you are right, they don't comute in general. I confused it with another situation, I'm sorry.

I have already read it in a book, but I didn't understand what it means. The most books say that all operators are time-independent in this picture. So, how can they have an explicit dependence on time?

I'm sorry for insisting on this question, I am really confused about pictures in quantum mechanics.
To be more specific, use as example a non-relativistic single particle. Then the fundamental set of operators are the position, momentum, and spin operators $\hat{\vec{x}}$, $\hat{\vec{p}}$, and $\hat{\vec{\sigma}}$, all of which by definition do not explicitly depend on time.

Now you can choose quite arbitrarily the picture of time evolution, i.e., how these fundamental observable operators depend on time by splitting the Hamiltonian $\hat{H}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{\sigma}},t)$ (where the time dependence here means explicit time dependence, e.g., for a charged particle moving in a time-dependent electromagnetic field) in two parts, $\hat{H}=\hat{H}_0+\hat{H}_1$.

Then the fundamental observable operators are defined to time-evolve with $\hat{H}_0$, i.e. according to the evolution equation
$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{\vec{x}}(t)=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}}(t),\hat{H_0}],$$
and analogously for $\hat{\vec{p}}$ and $\hat{\vec{\sigma}}$, while the state kets time-evolve according to $\hat{H}_1$:
$$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d}t} |\psi(t) \rangle= \hat{H}_1 |\psi(t) \rangle.$$
This is a general picture of time evolution (also knows as a "Dirac picture").

The Schrödinger picture lumps all time dependence to the states, i.e., you have $\hat{H}_0=0$ and $\hat{H}_1=\hat{H}$, while the Heisenberg picture lumps all time dependence to the operators representing observables, i.e., $\hat{H}_0=\hat{H}$ and $\hat{H}_1=0$.

Now for explicitly time dependent observable operators you have
$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{A}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{\sigma}},t) = \frac{1}{\hbar} [\hat{A},\hat{H}_0] + \partial_t \hat{A}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{\sigma}},t),$$
where the partial time derivative is due to the explicit time dependence and the commutator due to the time dependence of the fundamental operators.

For a general state, i.e., a statistical operator, on the one hand one has
$$\hat{\rho}=\sum_{j} P_j |\psi_j(t) \rangle \langle \psi_j(t)|,$$
and thus due to the equation of motion for the state kets
$$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \hat{\rho}= [\hat{H}_1,\hat{\rho}].$$
On the other hand, if read as a function of the fundamental operators (and mabe explicitly on time), it must fulfill
$$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \hat{\rho}=[\hat{\rho},H_0]+\mathrm{i} \hbar \partial_t \hat{\rho}.$$
Subtracting both equations yields the famous von Neumann equation
$$\mathring{\hat{\rho}}=\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}] + \partial_t \hat{\rho}=0.$$
Indeed
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}]+\partial_t \hat{A}$$
defines the operator for the time derivative $\dot{A}$ of the observable $A$, represented b the operator $\hat{A}$, i.e., that defines the "covariant time-derivative" in QT, and the von Neumann equation for the stat. op. states that the covariant time-derivative vanishes, as it must be in statistical mechanics.

• odietrich

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